The dipole moment of chlorobenzene is 1.5 D. | vedclass

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(C) The dipole moment of a molecule is the vector sum of the individual bond dipoles. For chlorobenzene, the dipole moment is $1.5 \ D$. Let the dipol

Vedclass · Accepted Answer

$2.54$

Vedclass · Answer

(C) The dipole moment of a molecule is the vector sum of the individual bond dipoles. For chlorobenzene, the dipole moment is $1.5 \ D$. Let the dipole moment of a $C-Cl$ bond be $\mu$. For $o-$dichlorobenzene, the two $C-Cl$ bonds are at an angle of $60^{\circ}$ to each other. The resultant dipole moment is given by $\mu_{net} = \sqrt{\mu^2 + \mu^2 + 2\mu^2 \cos(60^{\circ})}$. Since $\cos(60^{\circ}) = 0.5$, we have $\mu_{net} = \sqrt{2\mu^2 + 2\mu^2(0.5)} = \sqrt{3\mu^2} = \mu \sqrt{3}$. Given $\mu = 1.5 \ D$, the dipole moment of $o-$dichlorobenzene is $1.5 	imes 1.732 \approx 2.54 \ D$.

The dipole moment of chlorobenzene is $1.5 \ D$. The dipole moment of $o-$dichlorobenzene is (in $D$)

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